Question

Solve for x. 2+1/x-3/x^2=0

Answer 1

divide thru by the lcd to get rid of the fraction
lcd = x^2
this leaves a quadratic equation to be solved

Answer 2

this is
2x^2 + x -3 = 0
which can be factorized

Answer 3

Divide through? Or multiply through?

Answer 4

2+(1)/(x)-(3)/(x^(2))=0

Find the LCD (least common denominator) of 2+(1)/(x)-(3)/(x^(2))+0.
Least common denominator: x^(2)

Multiply each term in the equation by x^(2) in order to remove all the denominators from the equation.
2*x^(2)+(1)/(x)*x^(2)-(3)/(x^(2))*x^(2)=0*x^(2)

Cancel the common factor of x in the denominator of the first term (1)/(x) and the second term x^(2).
2*x^(2)+(1)/(<X>x<x>)*x^(<X>2<x>)-(3)/(x^(2))*x^(2)=0*x^(2)

Reduce the expression by removing the common factor of x in the denominator of the first term (1)/(x) and the second term x^(2).
2*x^(2)+1*x-(3)/(x^(2))*x^(2)=0*x^(2)

Cancel the common factor of x^(2) in the denominator of the first term -(3)/(x^(2)) and the second term x^(2).
2*x^(2)+1*x-(3)/(x^(<X>2<x>^(0)))*x^(<X>2<x>^(0))=0*x^(2)

Reduce the expression by removing the common factor of x^(2) in the denominator of the first term -(3)/(x^(2)) and the second term x^(2).
2*x^(2)+1*x-3*1=0*x^(2)

Multiply 2 by x^(2) to get 2x^(2).
2x^(2)+1*x-3*1=0*x^(2)

Multiply 1 by x to get x.
2x^(2)+x-3*1=0*x^(2)

Multiply -3 by 1 to get -3.
2x^(2)+x-3=0*x^(2)

Multiply 0 by x^(2) to get 0.
2x^(2)+x-3=0

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-6) that add up to b (1).In this problem (3)/(2)*-1=-(3)/(2) (which is (c)/(a)) and (3)/(2)-1=(1)/(2) (which is ((b)/(a)) , so insert (3)/(2) as the right hand term of one factor and -1 as the right-hand term of the other factor.
(x+(3)/(2))(x-1)=0

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(2x+3)(x-1)=0

Set each of the factors of the left-hand side of the equation equal to 0.
2x+3=0_x-1=0

Since 3 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 3 from both sides.
2x=-3_x-1=0

Divide each term in the equation by 2.
(2x)/(2)=-(3)/(2)_x-1=0

Cancel the common factor of 2 in (2x)/(2).
(<X>2<x>x)/(<X>2<x>)=-(3)/(2)_x-1=0

Remove the common factors that were cancelled out.
x=-(3)/(2)_x-1=0

Set each of the factors of the left-hand side of the equation equal to 0.
x=-(3)/(2)_x-1=0

Since -1 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 1 to both sides.
x=-(3)/(2)_x=1

The complete solution is the set of the individual solutions.
x=-(3)/(2),1

Answer 5

\[2 + 1/x-3/x ^{2}=0\]
Multiply all terms by x^2
\[2x ^{2}+x-3=0\] Factor
(x-1)(2x+3)=0 solve by setting each factor to 0
x-1=0, x=1
2x+3=0
2x=-3
x=-3/2
This is the same solution as above (but abbreviated)

Answer 6

Now I AM GOING TO GIVE A SIMPLE SULOTION (the above is too long).
Here: Make 2 + 1/x -3/x^2 = 0 to (by multiplying x^2)
2x^2 + x -3 =0/(1/x^2)=0
Then factorize it to:(x-1)(2x+3)=0
so x-1=0 and 2x+3=0
Then x = 1 or x=-3/2